# Discrepancy in inner product between tensor products

I have noticed one identity in case of tensor product from this post. But I can't understand why it is true.

$$\langle v_i| \otimes \langle w_j| \cdot |w_k\rangle \otimes |v_m\rangle = \langle v_i|v_m\rangle \cdot \langle w_j|w_k\rangle$$

Now if I consider $$\langle v_i|$$ to be $$[a,b]$$ and $$\langle w_j|$$ to be $$[c,d]$$. (note these are $$1 \times 2$$) then

$$\langle v_i| \otimes \langle w_j| = [a,b] \otimes [c,d] = [ac,ad,bc,bd]$$

Similarly $$|w_k\rangle \otimes |v_m\rangle = [e,f]^T \otimes [g,h]^T = [eg, eh, fg, fh]^T$$

Scalar product between them = $$[ac,ad,bc,bd] . [eg, eh, fg, fh]^T = aceg+ adeh+ bcfg+ bdfh$$

Now as per the identity if we do

$$\langle v_i|v_m\rangle = [a,b] . [g,h]^T = ag+bh$$ and $$\langle w_j|w_k\rangle = [c,d] . [e,f]^T = ce+df$$

So $$\langle v_i|v_m\rangle \cdot \langle w_j|w_k\rangle = (ag+bh).(ce+df)= aceg + adfg + bceh + bdfh$$

Now they are not the same. What is the reason for this behavior?

• Shouldn't ⟨v| = [a,b], ⟨vi| = a, ⟨vm| = b instead of ⟨vi| = [a,b] Nov 29, 2020 at 7:25
• @AbhayAravinda They do not denote the elements of the entry. Check the answer again. Nov 29, 2020 at 8:22

The identity is wrong. It should be $$\langle v_i | \otimes \langle w_j | \cdot | w_k \rangle \otimes | v_m \rangle = \langle v_i | w_k \rangle \langle w_j | v_m \rangle.$$
• In this case the more convenient formula is $\langle v_i| \otimes \langle w_j| \cdot |v_m\rangle \otimes |w_k\rangle = \langle v_i|v_m\rangle \cdot \langle w_j|w_k\rangle$. Nov 29, 2020 at 12:42